Implement atoi to convert a string to an integer.

Hint: Carefully consider all possible input cases. If you want a challenge, please do not see below and ask yourself what are the possible input cases.

Notes: It is intended for this problem to be specified vaguely (ie, no given input specs). You are responsible to gather all the input requirements up front.

Update (2015-02-10):
The signature of the C++ function had been updated. If you still see your function signature accepts a const char * argument, please click the reload button  to reset your code definition.

The function first discards as many whitespace characters as necessary until the first non-whitespace character is found. Then, starting from this character, takes an optional initial plus or minus sign followed by as many numerical digits as possible, and interprets them as a numerical value.

The string can contain additional characters after those that form the integral number, which are ignored and have no effect on the behavior of this function.

If the first sequence of non-whitespace characters in str is not a valid integral number, or if no such sequence exists because either str is empty or it contains only whitespace characters, no conversion is performed.

If no valid conversion could be performed, a zero value is returned. If the correct value is out of the range of representable values, INT_MAX (2147483647) or INT_MIN (-2147483648) is returned.

字串轉為整數是很常用的一個函式，由於輸入的是字串，所以需要考慮的情況有很多種。我之前有一篇文章是關於驗證一個字串是否為數字的，參見 http://www.cnblogs.com/grandyang/p/4084408.html 。在那篇文章中，詳細的討論了各種情況，包括符號，自然數，小數點的出現位置，判斷他們是否是數字。個人以為這道題也應該有這麼多種情況。但是這題只需要考慮數字和符號的情況：

1. 若字串開頭是空格，則跳過所有空格，到第一個非空格字元，如果沒有，則返回0.

2. 若第一個非空格字元是符號+/-，則標記sign的真假，這道題還有個侷限性，那就是在c++裡面，+-1和-+1都是認可的，都是-1，而在此題裡，則會返回0.

3. 若下一個字元不是數字，則返回0. 完全不考慮小數點和自然數的情況，不過這樣也好，起碼省事了不少。

4. 如果下一個字元是數字，則轉為整形存下來，若接下來再有非數字出現，則返回目前的結果。

5. 還需要考慮邊界問題，如果超過了整形數的範圍，則用邊界值替代當前值。

C++ 解法：

class Solution {
public:
    int myAtoi(string str) {
        if (str.empty()) return 0;
        int sign = 1, base = 0, i = 0, n = str.size();
        while (i < n && str[i] == ' ') ++i;
        if (str[i] == '+' || str[i] == '-') {
            sign = (str[i++] == '+') ? 1 : -1;
        }
        while (i < n && str[i] >= '0' && str[i] <= '9') {
            if (base > INT_MAX / 10 || (base == INT_MAX / 10 && str[i] - '0' > 7)) {
                return (sign == 1) ? INT_MAX : INT_MIN;
            }
            base = 10 * base + (str[i++] - '0');
        }
        return base * sign;
    }
};

Java 解法：

public class Solution {
    public int myAtoi(String str) {
        if (str.isEmpty()) return 0;
        int sign = 1, base = 0, i = 0, n = str.length();
        while (i < n && str.charAt(i) == ' ') ++i;
        if (str.charAt(i) == '+' || str.charAt(i) == '-') {
            sign = (str.charAt(i++) == '+') ? 1 : -1;
        }
        while (i < n && str.charAt(i) >= '0' && str.charAt(i) <= '9') {
            if (base > Integer.MAX_VALUE / 10 || (base == Integer.MAX_VALUE / 10 && str.charAt(i) - '0' > 7)) {
                return (sign == 1) ? Integer.MAX_VALUE : Integer.MIN_VALUE;
            }
            base = 10 * base + (str.charAt(i++) - '0');
        }
        return base * sign;
    }
}

參考資料：